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	<title>Comments on: September solutions are here!</title>
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		<title>By: Malcolm</title>
		<link>http://www.mathsolympiad.org.nz/2009/11/september-solutions-are-here/comment-page-1/#comment-191</link>
		<dc:creator>Malcolm</dc:creator>
		<pubDate>Fri, 20 Nov 2009 08:28:06 +0000</pubDate>
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		<description>There is also a nice inversive solution for problem 5 from the senior division I found a few days ago:

Invert about the point A, with arbitrary radius. You should end up with a picture that looks the same as the original one, but flipped, i.e. similar but oppositely oriented. Then, from the co-linearity of A,Q,(image of Q) and A,P,(image of P) and the fact that (image of P) corresponds to Q and (image of Q) corresponds to P under the similarity, angle PAO = angle QAO</description>
		<content:encoded><![CDATA[<p>There is also a nice inversive solution for problem 5 from the senior division I found a few days ago:</p>
<p>Invert about the point A, with arbitrary radius. You should end up with a picture that looks the same as the original one, but flipped, i.e. similar but oppositely oriented. Then, from the co-linearity of A,Q,(image of Q) and A,P,(image of P) and the fact that (image of P) corresponds to Q and (image of Q) corresponds to P under the similarity, angle PAO = angle QAO</p>
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